One usually sees $f(x):=\exp\frac{-1}{x^2}$ as an example of a $C^\infty$ function that is not analytic, having one point of non-analyticity (the point $0$).
The Fabius function is a canonical example of a $C^\infty$ function that is non-analytic on a continuum.
Consider now the (real) function $f(x)=\exp\frac{-1}{x^2}$ from above. With the understanding that $f$ is a bounded function and all derivatives of $f$ are bounded, define
$$g(x):=\sum_n 2^{-n}\ f(x-a_n)$$
Where $a_n$ is an enumeration of $\mathbb Q$. We get again a $C^\infty$ function, as uniform convergence of the sum and of the sum of derivatives follows from all derivatives (of $f$) being bounded.
It looks like $g$ is also nowhere analytic, since the points of non-analyticity of all the summands together is $\mathbb Q$, which is dense in $\mathbb R$ (if a function is analytic at $p$, there exists an open neighbourhood of $p$ on which it is also analytic).
But a proof is something different, and maybe, since we are putting non-analyticities arbitrarily close together, the non-analytic parts cancel at some points.
Is $g$ nowhere analytic?